 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/30j/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

w_{i}^{(1)} = 1/ n

. For

m = 1, \dots, M

choose $h_{m} \in B$ minimizing the weighted classification error
$ε_{m} = \sum_{i = 1}^{n} w_{i}^{(m)} 1_{{Y_{i} \neq = h_{m} (X_{i})}};$
(1)
set
$β_{m} = \frac{1}{2} lo g \frac{1 - ε _{m}}{ε _{m}};$
(2)
update and normalize
$w_{i}^{(m + 1)} = \frac{w _{i}^{(m)} e ^{- β_{m} Y_{i} h_{m} (X_{i})}}{\sum _{j} w _{j}^{(m)} e ^{- β_{m} Y_{j} h_{m} (X_{j})}} .$
(3)

Because

h \in B

implies

- h \in B

, the selected classifier can always have weighted error at most

1/2

, so

β_{m} \geq 0

. The output score is

f = m = 1 \sum M β_{m} h_{m} .

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